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If Sin θ + Cos θ = X, Prove that `Sin^6 Theta + Cos^6 Theta = (4- 3(X^2 - 1)^2)/4` - Mathematics

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प्रश्न

If sin θ + cos θ = x, prove that  `sin^6 theta + cos^6 theta = (4- 3(x^2 - 1)^2)/4`

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उत्तर

Given `sin theta + cos theta = x`

Squaring the given equation, we have

`(sin theta + cos theta)^2 = x^2`

`=> sin^2 theta + 2 sin theta cos theta = cos^2 theta = x^2`                                   

`=> (sin^2 theta + cos^2 theta) + 2sin theta cos theta = x^2`

`=> 1 + 2 sin theta cos theta = x^2`

`=> 2 sin theta cos theta = x^2  -1`

`=> sin theta cos theta = (x^2- 1)/2`

Squaring the last equation, we have

`(sin theta cos theta)^2 = (x^2  - 1)^2/4` 

`=> sin^2 theta cos^2 theta = (x^2 - 1)^2/4`

`=> sin^2 theta cos^2 theta = (s^2 -1)/4`

Therefore, we have

`sin^6 theta + cos^6 theta = (sin^2 theta)^3 + (cos^2 theta)^3`

`= (sin^2 theta + cos^2 theta)^3 -  3sin^3 theta cos^2 theta (sin^2 theta + cos^2 theta)`

`= (1)^3 - 3 ((x^2 - 1)^2)/4 (1)`

`= 1 - 3 (x^2 - 1)^2/4 (1)`

`x = 1 - 3 (x^2 - 1)^2/4`

`= (4- 3(x^2 - 1)^2)/4`

hence Proved

 

 

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पाठ 11: Trigonometric Identities - Exercise 11.1 [पृष्ठ ४७]

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आरडी शर्मा Mathematics [English] Class 10
पाठ 11 Trigonometric Identities
Exercise 11.1 | Q 86 | पृष्ठ ४७

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